A body is executing linear S.H.M. At a position x, its potential energy is ${\mathrm{E}}_1$, and at a position y, its potential energy is ${\mathrm{E}}_2$. The potential energy at the position (x + y) is

1.  ${\mathrm{E}}_1<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>{\mathrm{E}}_2$

2.  $\sqrt{{\mathrm{E}}_1^2<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>{\mathrm{E}}_2^2}$

3.  ${\mathrm{E}}_1<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>{\mathrm{E}}_2<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>2\sqrt{{\mathrm{E}}_1{\mathrm{E}}_2}$

4.  $\sqrt{{\mathrm{E}}_1{\mathrm{E}}_2}$

The equation of a particle executing simple harmonic motion is $y<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>0.4<mspace\; width="1em"></mspace>\sin \left(2\mathrm{\pi t}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>\frac{\mathrm{\pi }}{3}\right)$ (where t is in seconds and y is in meters). The initial phase of the particle is:

1.  $2\mathrm{\pi t}<mspace\; width="1em"></mspace>+<mspace\; width="1em"></mspace>\frac{\mathrm{\pi }}{3}$

2.  $\frac{\mathrm{\pi }}{3}$

3.  $\frac{7\mathrm{\pi }}{3}$

4.  $\frac{8\mathrm{\pi }}{3}$

An ideal spring-mass system has a time period of vibration T. If the spring is cut into 4 identical parts and same mass oscillates with one of these parts, then the new time period of vibration will be

1.  $\frac{\mathrm{T}}{2}$

2.  T

3.  $\frac{\mathrm{T}}{4}$

4.  2T

The equation of a particle executing simple harmonic motion is $y=2\sqrt{2}\sin 314t.$ Displacement y from the mean position where acceleration becomes zero is: (y is in cm and t is in second) 

1.  2 cm

2.  0

3.  $\frac{1}{\sqrt{2}}$ cm

4.  $2\sqrt{2}$ cm

The displacement \((\mathrm{x})\) of an SHM varies with time \((\mathrm{t})\) as shown in the figure. The frequency of variation of potential energy is:

                    
1. \(5\text{ Hz}\)
2. \(10\text{ Hz}\)
3. \(40\text{ Hz}\)
4. \(20\text{ Hz}\)

A simple pendulum bob is a hollow sphere full of sand suspended by means of a wire. If all the sand is drained out immediately, then the time period of the pendulum will:

A simple pendulum of length L is suspended from the ceiling of a cart which is sliding without friction on an inclined plane of inclination $\mathrm{\theta }$. The time period of the pendulum is

1.  $\mathrm{T}<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{\mathrm{g}}}$

2.  $\mathrm{T}<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{\mathrm{g}<mspace\; width="1em"></mspace>\mathrm{cos\theta }}}$

3.  $\mathrm{T}<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{\mathrm{g}<mspace\; width="1em"></mspace>\mathrm{sin\theta }}}$

4.  $\mathrm{T}<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{\mathrm{g}<mspace\; width="1em"></mspace>\mathrm{tan\theta }}}$

A body is placed on a horizontal platform that is undergoing vertical SHM. If the amplitude of oscillation is \(40\ \text{cm}\), then the least period of oscillation for which an object placed on the platform is not detached from it is:

1.  \(1.256\ \text{s}\)

2.  \(12.56 \ \text{s}\)

3.  \(0.1256 \ \text{s}\)

4.  \(125.6 \ \text{s}\)

If a pendulum giving correct time on the ground at a certain place is moved to the top of a tower 320 m high, then the loss in time (in seconds) measured by the pendulum clock in one week is:

1.  15.12

2.  7.72

3.  3.78

4.  30.24

Instantaneous acceleration (in ${\mathrm{ms}}^{-2}$) of a particle executing S.H.M. is given by $\mathrm{a}<mspace\; width="1em"></mspace>=<mspace\; width="1em"></mspace>-\frac{\mathrm{\pi }}{4}\sin \left(\frac{\mathrm{\pi }}{4}\mathrm{t}<mspace\; width="1em"></mspace>-<mspace\; width="1em"></mspace>\frac{\mathrm{\pi }}{6}\right)$.  The maximum speed of the particle will occur first time at

1.  1.75 s

2.  1.4 s

3.  1.2 s

4.  0.67 s